# Question #60c57

Dec 3, 2017

Q.7. Locomotive requires to exert constant force of $68 N$ per tonne when it is in motion. This force is required to overcome force of friction.

Force required to be exerted$= 30 \times 68 = 2040 N$

Additional force is required to accelerate the train. We know that rate of change of speed is acceleration. Using Newton's Second Law of motion and inserting values in SI units we get

$F = m a$
$F = 30000 \times 0.5 = 15000 N$

Total force required $= 2040 + 15000 = 17040$

$17.04 k N$

Dec 3, 2017

Q. 8. We know that

$\text{Power" ="work done"/"time taken}$ ......(1)

Work done $W$ in raising weight of $30 \text{ tonnes}$ through a distance of $24 m$

$W = \text{Force"xx"Distance}$ .....(2)

Also

$\text{Force} = m g$ ......(3)
where $m$ is mass of object and $g$ is acceleration due to gravity, $= 9.8 m {s}^{-} 2$

Inserting values in equation (1) in SI units we get

$\text{Power} = \frac{30000 \times 9.8 \times 24}{0.5 \times 60 \times 60}$
$\text{Power} = \frac{7056000}{1800} = 3920 W$

$3.92 k W$

Dec 3, 2017

Q. 9. Centrifugal force ${F}_{c}$ experienced by a body of mass $m$ while rotating in a circle of radius $r$ with angular velocity $\omega$ is given by the expression

${F}_{c} = m r {\omega}^{2}$ ......(1)

It is given that mass revolves about a point 2 meters from its centre of gravity. $\implies r = 2 m$

Units of $\omega \text{ are radians per second}$

Mass revolves at a speed of 60 revolutions per minute.
$\implies$ it revolves at a speed of 1 revolution per second.
One revolution is equal to $2 \pi$ radians
Hence $\omega = 2 \pi \text{ radians per second}$

Inserting values in (1) we get

${F}_{c} = 5 \times 2 \times {\left(2 \pi\right)}^{2}$

$\implies {F}_{c} = 394.8 N$, rounded to one decimal place.

Dec 3, 2017

Q. 10. See solution here.